Question

consider the equation and the given point.
$f(x) = (x^3 + 7x - 1)(x - 2), \\ (1, -7)$
(a) find an equation of the tangent line to the graph of $f$ at the given point.
$y = $
(b) use a graphing utility to graph the function and its tangent line at

Explanation:

Step1: Expand the function

First, multiply out the terms of $f(x)$:

$$\begin{align*} f(x)&=(x^3 + 7x - 1)(x - 2)\\ &=x^3(x-2)+7x(x-2)-1(x-2)\\ &=x^4 - 2x^3 + 7x^2 - 14x - x + 2\\ &=x^4 - 2x^3 + 7x^2 - 15x + 2 \end{align*}$$

Step2: Find the derivative $f'(x)$

Use the power rule $\frac{d}{dx}[x^n]=nx^{n-1}$:

$$ f'(x)=4x^3 - 6x^2 + 14x - 15 $$

Step3: Calculate slope at $x=1$

Substitute $x=1$ into $f'(x)$ to get the tangent slope:

$$\begin{align*} f'(1)&=4(1)^3 - 6(1)^2 + 14(1) - 15\\ &=4 - 6 + 14 - 15\\ &=-3 \end{align*}$$

Step4: Use point-slope form

Point-slope formula: $y - y_1 = m(x - x_1)$, where $(x_1,y_1)=(1,-7)$ and $m=-3$:

$$\begin{align*} y - (-7)&=-3(x - 1)\\ y + 7&=-3x + 3\\ y&=-3x - 4 \end{align*}$$

Answer:

$y=-3x-4$

Note: Part (b) requires a graphing utility, which cannot be completed in this text format, but the equation for the tangent line needed for graphing is provided above.